Biomedical Engineering Reference
In-Depth Information
, a linear form of (7.23) is used where ln(
X
/
X
0
) is
plotted against time and the slope corresponds to
To determine the value of
μ
μ
. A smaller
μ
value indicates that
the cell cycle is longer.
EXAMPLE 7.4
To understand the effect of newly created material, Stella seeded 10,000 cells in a tissue
culture plate. To measure the growth rate, she added
3
H thymidine, a nontoxic radioac-
tive substance, to the culture and read the incorporated radiation (counts per minute) 24
hours later. The radiation reading was 5,000 counts per minute (cpm) which correlates
to the amount of DNA synthesized.
To calibrate the amount of cpm to number of cells,
she recounted the cells after 24 hours and found that the total cell number is 20,000 per
culture plate. If she adds a specific soluble factor to the medium under the same condi-
tions (10,000 cells/well over 24 hours), the incorporated radiation reads 8,000 cpm. What
is the specific growth rate constant under the new conditions? List few assumptions used
during these calculations.
Solution:
X
0
is 10,000 cells and
X
(
t
= 24 hours): 20,000 cells
Hence, a growth rate of 10,000 new cells/day corresponds to 5,000 cpm (i.e., 0.5
cpm/cells in 24 hours).
If the cpm count is 8,000,
Number of newly proliferated cells = 8,000/0.5 =16,000 cells
Hence,
X
(
t
= 24) =
X
0
+ 16,000 = 26,000
From (7.23), 26,000 = 10,000 exp(24
μ
)
Hence,
μ
=
0.040 hr
−
1
The assumptions are that the growth rate is directly proportional to
3
H thymidine
uptake and is taken up only during cell division and is in excess so that it will not run out.
The plating efficiency is 100% (i.e., all the 10,000 cells are capable of dividing).
cpm = 0 when there is no proliferation (i.e., the background signal is negligible).
For comparison purposes, the time interval
t
d
required to double the popula-
tion is sometimes calculated. Nevertheless, (7.23) is applicable for cells in the expo-
nential growth phase. A lag phase or time required to adjust to a new environment
can be included by correcting time, that is,
XX
=
0
exp( (
μ
tt
−
))
(7.25)
lag
However, if only a fraction (
f
) of the cells are dividing and cells have different
phases, then the total number of cells after one cell cycle is given by
Xf XfX
=−
(1
)
+
2
(7.26)
0
0
